NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.3 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths chapter 6 Permutations and Combinations Ex 6.3.
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.3
Ex 6.3 Class 11 Maths Question 1.
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Solution.
Total number of digits are 9. We have to form 3-digit numbers without repetition.
∴ The required number of 3-digit numbers = 9P3
= 9!/6! = 9 × 8 × 7 = 504
Ex 6.3 Class 11 Maths Question 2.
How many 4-digit numbers are there with no digit repeated?
Solution.
The 4-digit numbers are formed from the digits 0 to 9. In 4-digit numbers 0 is not taken at the thousands place, so the thousands place can be filled in 9 different ways. After filling thousands place, 9 digits are left. The remaining three places can be filled in 9P3 ways.
So, the required number of 4-digit numbers
= 9 × 9P3
= 9 × 504 = 4536
Ex 6.3 Class 11 Maths Question 3.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution.
For 3-digit even numbers, unit place can be filled by 2, 4, 6, i.e., in 3 ways. Then, the remaining two places can be filled in 5P2 ways.
∴ The required number of 3-digit even numbers
= 3 × 5P2
= 60
Ex 6.3 Class 11 Maths Question 4.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Solution.
The 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 in 5P4 ways.
∴ The required number of 4-digit numbers = 5P4 = 120
For 4-digit even numbers unit place can be filled by 2, 4, i.e., in 2 ways. Then, the remaining three places can be filled in 4P3 ways.
∴ The required number of 4-digit even numbers
= 2 × 4P3 = 2 × 24 = 48
Ex 6.3 Class 11 Maths Question 5.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
Solution.
From a committee of 8 persons, we can choose a chairman and a vice chairman in 8P2 ways.
Thus, the number of ways we can choose a chairman and vice chairman assuming one person cannot hold more than one position = 8P2 = 56.
Ex 6.3 Class 11 Maths Question 6.
Find n if n-1P3 : nP4 = 1 : 9.
Solution.
Ex 6.3 Class 11 Maths Question 7.
Find r if
(i) 5Pr = 2 6Pr-1
(ii) 5Pr = 6Pr-1
Solution.
Ex 6.3 Class 11 Maths Question 8.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Solution.
The number of letters in the word EQUATION = 8
∴ The number of words that can be formed
= 8P8 = 8!
= 40320
Ex 6.3 Class 11 Maths Question 9.
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
Solution.
The number of letters in the word MONDAY = 6
(i) When 4 letters are used at a time.
Then, the required number of words made = 6P4
= 6!/2! = 6 × 5 × 4 × 3 = 360
(ii) When all the letters are used at a time.
Then, the required number of words make = 6P6 = 6!
= 720
(iii) When all the letters are used but first letter is a vowel.
So, the first letter can be either A or O.
Thus, there are 2 ways to fill the first letter and remaining places can be filled in 5P5 ways.
∴ The required number of words made = 2 × 5P5
= 2 × 5! = 240
Ex 6.3 Class 11 Maths Question 10.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Solution.
There are 11 letters in the word MISSISSIPPI, of which I appears 4 times, S appears 4 times, P appears 2 times and M appears 1 time.
∴ The required number of arrangements
= 11!/(4!4!2!) = (11 × 10 × 9 × 8 × 7 × 6 × 5)/(4 × 3 × 2 × 1 × 2 × 1)
= 34650 … (i)
When four I’s come together, we treat them as a single letter. So, the total number of letters in the word will be 8. These 8 letters in which there are four S’s and two P’s
can be rearranged in 8!/4!2! ways, i.e., in 840 ways. … (ii)
Hence, the number of arrangements when four I’s do not come together
= 34650 – 840 = 33810
Ex 6.3 Class 11 Maths Question 11.
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Solution.
There are 12 letters in the word PERMUTATIONS of which T appears 2 times.
(i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times.
∴ The required number of words = 10!/2!
= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 1814400
(ii) When vowels are taken together, i.e., E U A I O we treat them as a single letter. This single letter with remaining 7 letters will account for 8 letters, in which there are two Ts, which can be rearranged in 8!/2! = 20160 ways.
Corresponding to each of these arrangements, the 5 vowels E, U, A, I, O can be rearranged in 5! ways = 120 ways. Therefore, by multiplication principle, the required number of arrangements = 20160 × 120 = 2419200.
(iii) When there are always 4 letters between P and S
∴ P and S can be at
1st and 6th place
2nd and 7th place
3rd and 8th place
4th and 9th place
5th and 10th place
6th and 11th place
7th and 12th place.
So, P and S will be placed in 7 ways and can be arranged in 7 × 2! ways = 14 ways
The remaining 10 letters with two T’s, can be arranged in 10!/2! ways = 1814400 ways.
∴ The required number of arrangements = 14 × 1814400 = 25401600.
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